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Sergeu [11.5K]
2 years ago
12

The electron configuration of an element is 1s22s22p63s1. Describe what most likely happens when an atom of this element comes n

ear an atom having seven valence electrons. Source stylesformat.
Chemistry
1 answer:
tia_tia [17]2 years ago
8 0

The element will lose it's valence electron (i.e 1 electron) to form ion when it comes close to an atom having 7 valence electrons.

The electronic configuration of the element is given below:

Electronic configuration => 1s² 2s²p⁶ 3s¹

From electronic configuration, we can see that the element has 1 valence electron.

This implies that the element will lose it's valence electron when it comes close to an atom having 7 valence electrons

This can further be seen as follow:

Electronic configuration of element => 1s² 2s²p⁶ 3s¹

Atomic number of element = 2 + 2 + 6 + 1 = 11

Name of element = sodium, Na

Element will 7 valence = chlorine, Cl

Na –> Na⁺ + e

Cl + e —> Cl¯

Na + Cl —> Na⁺ Cl¯

From the above illustration, we can see that the element (i.e Na) loses it's valence electron to form Na⁺ when it come close to Cl (having 7 valence electrons)

See attachment photo for further details

Learn more on group 1 and 7 elements: brainly.com/question/337107

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3.65 gram of hcl is dissolved in 180 gram of water. Find the total number of molecules of hydrogen​
Morgarella [4.7K]

Answer:

Molec_{\ H_{tot}}=1.206x10^{25}molec

Explanation:

Hello.

In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:

molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH}  =6.03x10^{22}molec

Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:

molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH}  =1.20x10^{25}molec

Thus, the total number of molecules turns out:

Molec_{\ H_{tot}}=6.03x10^{22}+1.20x10^{25}\\\\Molec_{\ H_{tot}}=1.206x10^{25}molec

Regards.

6 0
3 years ago
Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow
Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

   Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

         \frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)

      \rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h

   \frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}

          [/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = \frac{10}{\rho A_{c}}

                       A_{c} = 0.01 m^{2}

              \frac{dh}{dt} + h = 1

                  \frac{dh}{dt} = 1 - h

               \frac{dh}{1 - h} = dt  

                \frac{ln(1 - h)}{-1} = t + C      

Given at t = 0 and V = 0  

                         A \times h = 0  

 or,                     h = 0

                 -ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

                                C = 0  

                So,   -ln(1 - h) = t

or,                      t = ln (\frac{1}{1 - h})  ........... (1)

(a)    Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

                    t = ln (\frac{1}{1 - h})  

                     t = ln (\frac{1}{1 - 0.6})  

                        = ln (\frac{1}{0.4})

                        = 0.916 seconds

(b)   As maximum height of water level in the tank is achieved at steady state that is, t = \infty.  

                    1 - h = exp (-t)

                    1 - h = 0  

                         h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

                 

8 0
3 years ago
A carbon molecule that has a different arrangement of atoms is known as a/an
maw [93]
The correct response I believe is A. Isomer. If a carbon molecule possess more than one arrangement of how its atoms can be arranged, those other arrangements are known as isomers.
3 0
3 years ago
4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a
yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0
2 years ago
What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
Gekata [30.6K]
This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
7 0
3 years ago
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